971

已知正数 ${\displaystyle a,b,c}$ 满足 ${\displaystyle {\mathrm{e}}^a={1.1}^3,5b^2+10b-3=0,{\mathrm{e}}^c=1.3}$, 比较 ${\displaystyle a,b,c}$ 大小.

由题意 ${\displaystyle a=3\ln 1.1,b={\displaystyle \frac{2\sqrt{10}}{5}-1,c=\ln 1.3}}$. 则一方面根据 ${\displaystyle \ln x>{\displaystyle \frac{2(x-1)}{x+1},x>1}}$, 有$$\begin{array}{r}a-b>3\frac{2(1.1-1)}{1.1+1}-b=\frac{9}{7}-\frac{2\sqrt{10}}{5}=\frac{\sqrt{2025}-\sqrt{1960}}{5}>0;\end{array}$$
另一方面在 ${\displaystyle \ln (x+1)={\displaystyle -\sum _{i=1}^{\mathrm{\infty }}\frac{(-x{)}^i}{i}}}$ 中取三项，构造不等式 ${\displaystyle {\displaystyle \ln (x+1)<x-\frac{x^2}{2}+\frac{x^3}{3}}}$,有 $$\begin{array}{r}c-b<0.3-\frac{{0.3}^2}{2}+\frac{{0.3}^3}{3}-b=\frac{158}{125}-\frac{2\sqrt{10}}{5}=\frac{2(\sqrt{6241}-\sqrt{6250})}{125}<0,\end{array}$$
从而 ${\displaystyle c<b<a}$.